Percent change = (new number - old number)/(old number) * 100
A positive percent change is a percent increase.
A negative percent change is a percent decrease.
In this problem, we have:
The new number is 15 laps.
The old number is 12 laps.
percent change = (15 - 12)/(12) * 100
percent change = 3/12 * 100
percent change = 25
Since the percent change is positive, +25, it is a percent increase.
Answer: The percent increase is 25%
Paul bakes 300 bread loaves. I don’t know the question you’re asking for the second question.
(a) 0.059582148 probability of exactly 3 defective out of 20
(b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.
(a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So
0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)
= 0.05^3 * 0.95^17 * 20! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)
= 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)
= 0.05^3 * 0.95^17 * 20*19*3
= 0.000125* 0.418120335 * 1140
= 0.059582148
(b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.
The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)
= 0.05*0.95^3*24/(1!3!)
= 0.05*0.857375*24/6
= 0.171475
So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.
Answer:
48
Step-by-step explanation:
0.8 times 60= 48
Answer:
The 95% confidence interval for the proportion of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694)
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of 
, and a confidence level of 
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of 
.
For this problem, we have that:
95% confidence level
So 
, z is the value of Z that has a pvalue of 
, so 
.
The lower limit of this interval is:
The upper limit of this interval is:
The 95% confidence interval for the proportion of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694)
